Problem: $\overline{BC} = 6$ $\overline{AC} = {?}$ $A$ $C$ $B$ $?$ $6$ $ \sin( \angle BAC ) = \frac{2\sqrt{5} }{5}, \cos( \angle BAC ) = \frac{ \sqrt{5}}{5}, \tan( \angle BAC ) = 2$
$\overline{BC}$ is the opposite to $\angle BAC$ $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the opposite side and need to solve for the adjacent side so we can use the tan function (TOA) $ \tan( \angle BAC ) = \frac{\text{opposite}}{\text{adjacent}} = \frac{\overline{BC}}{\overline{AC}}= \frac{6}{\overline{AC}} $ $ \overline{AC}=\frac{6}{\tan( \angle BAC )} = \frac{6}{2} = 3$